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Let \(f(t)\) be a piecewise continuous function on \([0,\infty )\). The Laplace transform of \(f\) is the function \(F(s)=\displaystyle \int _0^{\infty } f(t)e^{-st}\,dt\), on the set of \(s\) values for which the integral converges.
If \(s=0\) then \(\displaystyle \int _0^{\infty } \cos {(\omega t)}e^{-st}\,dt=\int _0^{\infty } \cos {(\omega t)}\,dt\), which ... .
If
\(s<0\) then \(\displaystyle \int _0^{\infty } \cos {(\omega t)}e^{-st}\,dt\) also ... . For \(s> 0\), let
\(\displaystyle A=\int _0^{\infty } \cos {(\omega t)}e^{-st}\,dt\). By ... with \(u=\cos {(\omega t)}\)
and ... we get
that
\(A=\) ... \(\displaystyle \left .\:^{\;}\right |_0^{\infty } - \frac {\omega }{s} \int _0^{\infty }\) ... \(dt\).
For \(s>0\), \(\displaystyle \lim _{t\rightarrow \infty } -\frac {\cos {(\omega t)}}{s}e^{-st}=\) ...
so \(\displaystyle A=\frac {1}{s} \; -\) ... .
By ... , with \(\displaystyle u=\sin {(\omega t)}\) and
... we get
\(\displaystyle A=\frac {1}{s} \; - \; \frac {\omega }{s} [\) ... \(]\).
For \(s>0\), \(\displaystyle \lim _{t\rightarrow \infty } -\frac {\sin {(\omega t)}}{s}e^{-st}=\) ... .
Then \(A=\frac {1}{s} \; - [\) ... \(]A\). Solving this for \(A\)
we get \(A=\) , Hence the
... of \(\cos (\omega t)\) is
\(A=\) ... for \(s>0\).